Physics optics light behavior question?

Question

A light source, S, is located 2.0m below the surface of a swimming pool and 1.5m from one edge of the pool. The pool is filled to the top with water.
a) At what angle does the light reaching the edge of the pool leave the water?
b) Does this cause the light viewed from this angle to appear deeper or shallower than it actually is?

Answer 1

Since the diagonal distance cannot be 1.5 m, this distance denotes the horizontal distance from the edge. It is usual in this type of problems to give x and y distances.
Vertical distance is 2m.
horizontal distance is 1.5 m
Hence tan θ = 2/1.5 => θ = 53.13
The ray is at angle 53.13° from the base of the pool and is incident on the edge of the pool. The incident angle is 90 - 53.13°=36. 87°
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The incident angle r can also be found directly since
tan r = 1.5 /2 and r =36. 87°
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μ for water is 1.33.
1.33 = sin i / sin r where r = 36. 87°
sin i = 1.33*sin r = 1.33 *0.6
i = 52.94
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Since i is less than 53.13
the ray appears shallower than it actually is.

Answer 2

Alto >>>
how can light source be vertically down =2m
and diagonally away = 1.5 m
it violates Pythagorous theoram
2^2 + x^2 = 1.5^2 ?????
have you got it interchanged

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Pearlsawme >>> knows all solved questions >> he has taken tangent by his own convenience?
what is the source of where r = 33.69 ° ???? from internet??